(I seem to be having trouble posting this reply; I suspect my attempts to type inequalities are being misconstrued as really bad HTML…)

Even more cool (to me) is the fact that there are no other approximations of this kind — since 2 is the only integer for which ln(2) is less than 1, it’s the only one that can generate a difference of ~ 1 between 1/ln(r) and 1/ln(k) for integers r and k.

Kim, I’m using the fact that logA(x) – logB(x) = ln(x)[ 1/ln(A) – 1/ln(B)]. If you want to approximate ln(x) using a difference of logs of other bases, you want to find integers A, B, and N such that N*[1/ln(A) – 1/ln(B)] is as close to 1 as possible. In the example above, 1/ln(15) – 1/ln(41) is very close to 0.1. Those values of A and B were found empirically.